An equipoterntial surface that surrounds a + 3.0 pC point charge has a radius of 2.0 cm. What is the potential of this surface?
1 answer:
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Answer:
1) The maximum height reached is approximately 39.715 m
2) The time it takes the ball to reach the its highest point is approximately 2.85 s
Explanation:
1) The vertical velocity of the ball, u = 27.9 m/s
The acceleration due gravity, g = 9.8 m/s²
The kinematic equation that gives the height, h, reached by the ball is given as follows;
v² = u² - 2·g·h
Where;
v = The final velocity of the ball = 0 m/s at the maximum height
h = The height reached by the ball = at maximum height
Therefore, by substitution of the known values, we have;
v² = u² - 2·g·h
0² = 27.9² - 2 × 9.8 ×
2 × 9.8 × = 27.9²
= 27.9²/(2×9.8) ≈ 39.715
The maximum height reached = ≈ 39.715 m
2) From the kinematic equation of motion, v = u - g·t, we have the time, t, it takes the ball to reach the maximum height given as follows
At maximum height, the final velocity, v = 0 m/s, therefore, we have;
0 = 27.9 - 9.8 × t
9.8 × t = 27.9
t = 27.9/9.8 ≈ 2.85
The time it takes the ball to reach the maximum height = t ≈ 2.85 seconds.
Answer:
1225 J
Explanation:
The Gravitational potential energy (PEG) gained by a mass lifted above the ground is given by
where
m is the mass
g = 9.8 m/s^2 is the acceleration due to gravity
h is the height at which the object has been lifted
In this problem, we have
m = 250 kg
h = 0.5 m
So, the PE of the object is